Analyzing functions #6 - Questions and Answers

Question 1 Write some examples of non differentiable functions Answer 1 There are some cases when a function can be non-differentiable. Firstly, a function in non-differentiable where it is discontinuous. For example, f(x) = cot x is non-differentiable at x = nπ for all integer n. Thus, the graph is {y = cot x [-10, 10, -5,5]} Another example: f(x) = (x³-6x²+9x) / (x³ - 2x² - 3x) is non-differentiable at 0 and at 3 and at -1. Note that f(x) = (x(x-3)²)/(x(x-3)(x+1)) And the graph: {(x³ - 6x² + 9x)/(x³ - 2x² - 3x) [-10, 10, ,-5,5]} If we define f(x) to be 0 if x is a rational number and 1 if x is irrational, the function is non-differentiable at all x. A function is non-differentiable where it has a "cusp" or a "corner point". If f'(x) is defined for all x near a (all x in an open interval containing a) except at a, but lim(x→a-) f'(x) ≠ lim(x→a+) f'(x). Another example: f(x) = |x-2|. It is non-differentiable at 2. (This function can also be written as f(x) = √(x² - 4x + 4) So, the graph: {|x-2| [-3.86, 10.184, -3.45, 3.57]} Example: f(x) = x + √(x²-2x + 1) Is non-differentiable at 1. Graph: x + √(x²-2x+1) [-3.86, 10.184, -3.45, 3.57] Finally, a function is non-differentiable at a if it has a vertical tangent line at a. For example: f(x) = 2 + (x-1)¹/³ has vertical tangent line at 1. And therefore is non- differentiable at 1. Graph: 2 + (x - 1)¹/³ [-2.44, 4.487, -0.353, 3.111]} For some functions, we only consider one-sided limts: f(x) = √(4 – x²) has a vertical tangent line at -2 and at 2. lim(x→2-) f'(x) does not exist, but lim(x→2-) |f'(x)| = ∞ Answer 2 Well, the most common forms of non-differentiable behavior involve a function going to infinity at x, or having a jump or cusp at x. However, the function sin(1/x) for example is singular at x = 0 even though it always lies between -1 and 1. Its hard to say what it does right near 0 but it sure doesn't look like a straight line. Question 2 Explain how to apply the First Derivative Test. Answer 1 The first derivative test is the process of analyzing functions using their first derivatives in order to find their extremum point. This process is called the First derivative test. Now if f'(x) changes positive to negative at the critical point then local maximum and if the derivative of the function changes negative to positive at the critical point the minimum and if f'(x) does not change signs then the function is neither maxima nor minima at that critical point. Answer 2 The second derivative test is used to find the relative maxima or relative minima. The second derivative test: Suppose the function f(x) has critical point at x = c that is f'(c) = 0 and f''(x) is continuous in the neighborhood of c, then If f''(c) < 0 then f(x) has relative minima at x = c. If f''(c) = 0 then f(c) can be relative minima or relative maxima or neither. Steps to apply the second derivative test: Find all critical points of function y= f(x) that is the points at which f'(x) = 0 or f'(x) does not exist. Compute the second derivative of f(x) at those critical points. a) If the second derivative at some critical point is positive then at that point the function f(x) has relative maxima. b) If the second derivative at some critical point is negative then at that point the function f(x) has relative minima. and c) If second derivative is zero at some critical point then test is inconclusive. Question 3 what are the inflection point relationship of stationary point and critical point what are the critical points f(θ) = sin θcos²θ - cotθ/θ + 1 Domain: [0, 2π] Answer What are inflection points? An inflection point is where the curve of the graph goes from concave down to up or vice versa. Point of inflection occur at f"=0 Relationship of stationary point and critical point. We say x₀ is a stationary point of a function if f(x) and f'(x) exist and is equal to f'(x₀)=0. And, x₀ is a critical point of a function of f(x) if f(x₀) exists and either f'(x₀) does not exist (i.e. function is not differentiable at or f'(x₀)=0. All stationary points are critical points but not all critical points are stationary points. What are the critical points? Point x₀ is a critical point of a function of f(x) if f(x₀) exists and either f'(x₀) does not exist (i.e. function is not differentiable at x₀) or f'(x₀)=0. f(θ) = sin θcos²θ - cot θ/θ + 1 Domain: [0, 2π] The first derivative test is a method of analyzing functions using their first derivatives in order to find their extremum point. We take the derivative of the function, f'(θ) = d/dθ (sin θcos²θ - cot θ/θ +1) = d/dθ (sin θcos²θ) - d/dθ (cot θ/θ) + d/dθ (1) = cos³θ + θcosec²θ - cotθ/θ² - sin2θ/sinθ Now, we substitute θ=4, f'(θ) = -0.27927 - (-0.49047) - (-0.74875) =0.959951 ≈ 0.9600 Question 4 How do I find concave up and concave down from f(x) = x³ + 3x² + 5x + 7? Answer Step 1 First find the derivative: f'(x) = 3x² + 6x + 5. Next find the second derivative: f''(x) = 6x + 6 = 6(x + 1). The second derivative changes sign from negative to positive as x increases through the value x = -1. Step 2 Therefore the graph of f is concave down when x < -1, concave up when x > -1, and has an inflection point when x = -1. The coordinates of the inflection point are (-1, f(-1)) = (-1, 10).